Is ${913231}$ divisible by $4$ ?
A number is divisible by $4$ if the last two digits are divisible by $4$ . [ Why? We can rewrite the number as a multiple of $100$ plus the last two digits: $ \gray{9132} {31} = \gray{9132} \gray{00} + {31} $ Because $913200$ is a multiple of $100$ , it is also a multiple of $4$ So as long as the value of the last two digits, ${31}$ , is divisible by $4$ , the original number must also be divisible by $4$ Is the value of the last two digits, $31$ , divisible by $4$ No, $31$ is not divisible by $4$, so $913231$ is also not divisible by $4$.